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What would be the co-efficient of $x^{60}$ in the expansion of $\space$ $\prod_{m=1}^{11} (x^m - m)$ ?

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4 Answers 4

HINT: Imagine multiplying out that product without collecting like terms. Each uncollected term has the form $$u_1u_2u_3\dots u_{11}\;,\tag{1}$$

where $u_k$ is either $x^k$ or $-k$. Let $S=\{k:u_k=x^k\}$; then the term $(1)$ is an $x^{60}$ term iff $\sum S=60$, and its coefficient is the product of the $-k$ for $k\notin S$.

Now $\sum_{k=1}^{11}k=\frac{11\cdot12}2=66$, so there aren’t really very many subsets of $\{1,\dots,11\}$ whose sum is $60$; just find all of them, calculate the corresponding terms of type $(1)$, and add them.

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Note that $(a + b)(c + d) = ac + ad + bc + bd$. Each term in the expansion is the product of one of the two summands in the first bracket and one of the two summands in the second bracket (and every such product appears in the expansion). This is also true when more factors are included i.e. $(a + b)(c + d)(e + f)$, etc.

In the expansion of $\prod_{m=1}^{11}(x^m - m)$, before collecting like terms, each term in the expansion is the product of one of the two summands from each bracket. Which means we need to determine what possible choices could be made to come up with an $x^{60}$ term. As $x^a.x^b = x^{a+b}$ we are looking for all possible products $x^{a_1}\dots x^{a_n}$ with $a_1, \dots, a_n \in \{1, \dots, 11\}$, $a_i \neq a_j$ for $i\neq j$ and $a_1 + \dots + a_n = 60$. That is, we need to determine the ways to obtain $60$ as a sum using each of the numbers $1, \dots, 11$ at most once. As $1 + \dots + 11 = 66$, it isn't too hard to write out all the possible ways to obtain $60$ in this way.

One example is $1 + 2 + 3 + 4 + 5 + 7 + 8 + 9 + 10 +11 = 60$. Then in the expansion of the product, this decomposition of $60$ corresponds to the term which is obtained as the product of $x^1$ (from the first bracket), $x^2$ (from the second bracket), $\dots$, $x^5$ (from the fifth bracket), $-6$ (from the sixth bracket), $x^7$ (from the seventh bracket), $\dots$, $x^{10}$ (from the tenth bracket), and $x^{11}$ (from the eleventh bracket). This is the term

$$x^1.x^2.x^3.x^4.x^5.(-6).x^7.x^8.x^9.x^{10}.x^{11} = -6x^{60}.$$

If you can figure out all the decompositions of $60$ as outlined above, and then calculate the corresponding element of the expansion of $\prod_{m=1}^{11}(x^m - m)$, you will obtain (not many) $x^{60}$ terms. Adding them up will give you the $x^{60}$ term in the final expansion at which point you will be able to read off the desired coefficient.


You can actually take a short cut. It is equivalent (convince yourself why) to find all possible ways to write $6$ as a sum using the numbers $1, \dots, 6$ at most once, and then for each sum $a_1 + \dots + a_k$, calculate $(-1)^ka_1\dots a_k$. Adding up the resulting numbers for each decomposition gives you the desired coefficient.

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The product $\space$ $\prod_{m=1}^{11} (x^m - m)$ is $$x^{66}-x^{65}-2x^{64}-x^{63}-x^{62}+5x^{61}+x^{60}+13x^{59}+4x^{58}+...$$ so the coefficient of $x^{60}$ is $1$. There may be a clever way to arrive at this, but since the coefficients seem to be a bit erratic, it seems perhaps it's just a coincidence that it comes out 1, due to some cancellations.

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The degree of the given polynomial is , $1+2+...+11=66$ . $66-60=6$ . We have to consider polynomials of degree $6$ ; $x^6-6$ , $(x^2-2)(x^4-4)$ , $(x-1)(x^5-5)$ , $(x-1)(x^2-2)(x^3-3)$ .

The coefficient of $x^{60}$ is the sum of the constant terms i.e., without $x$ , terms in these $4$ polynomials, that is $\space$ $-6+8+5-6=1$

$HAPPY$ $NEW$ $YEAR$$\space$ $!$

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Why do polys of degree only 6 have to do with the degree 60 term? Are you using a symmetry of the polynomial? If so please add to your answer. –  coffeemath Jan 3 '13 at 6:08

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